Question
$$\sum\limits_{r = 1}^n {{{\sin }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right)} \,$$ is equal to
A.
$${\tan ^{ - 1}}\left( {\sqrt n } \right) - \frac{\pi }{4}$$
B.
$${\tan ^{ - 1}}\left( {\sqrt {n + 1} } \right) - \frac{\pi }{4}$$
C.
$${\tan ^{ - 1}}\left( {\sqrt n } \right) $$
D.
$${\tan ^{ - 1}}\left( {\sqrt {n + 1}} \right) $$
Answer :
$${\tan ^{ - 1}}\left( {\sqrt n } \right) $$
Solution :
$$\eqalign{
& {\sin ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right) = {\tan ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{1 + \sqrt r \sqrt {\left( {r - 1} \right)} }}} \right) \cr
& = {\tan ^{ - 1}}\sqrt r - {\tan ^{ - 1}}\left( {\sqrt {r - 1} } \right) \cr
& \Rightarrow \sum\limits_{r = 1}^n {{{\sin }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {r - 1} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right)} \cr
& = \sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}\sqrt r - {{\tan }^{ - 1}}\sqrt {r - 1} } \right)} = {\tan ^{ - 1}}\sqrt n \cr} $$